Show only one group exists for Groups of prime order

I'm trying to dust off the cobb webs by studying some basic Group Theory. Can someone provide a proof that a Group of order 5, or any prime for that matter, must be a cyclic Abelian group, and that there can be only one such group...

I can easily find the multiplication table for the Group, but I don't see how to prove the statement that it is the only such group.

You might try thinking about Lagrange's Theorem, which states that if is a finite group, the order of any subgroup must divide the order of . Can you see how to use this to show that every group of prime order is cyclic?

You might try thinking about Lagrange's Theorem, which states that if is a finite group, the order of any subgroup must divide the order of . Can you see how to use this to show that every group of prime order is cyclic?

And of course then use this and the important fact that every cyclic group is isomorphic to or to finish.

You might try thinking about Lagrange's Theorem, which states that if is a finite group, the order of any subgroup must divide the order of . Can you see how to use this to show that every group of prime order is cyclic?

I see that a group of prime order cannot have any subgroups (other than E), but how does that lead to the non-existance of groups of a given prime order, other than the cyclic, Abelian group? For example, I can create more than one multiplication table for which obeys the rule that no element appear more than once in any row or column. However, I can show the table for the non-Abelian violates the associative property, and thus is not a proper group. But such a brute force approach is not practical for primes of higher order.